You ask how he figures that and he says, "I have $100 from the US Treasury and I printed $900 on my ink jet printer this morning." We hope he only gets 9 out of the 10 years he would normally serve. At the bottom of the curves, where the triode should cutoff, the grid lines of the spongy tube are almost horizontal, which means the rp is no longer a low 280 ohms, but an extremely high 5k. This means that the tube no longer fights any movement on its plate as it now resembles a current source. Down here, the tube fails to provide grip on the output signal and is just a soft spongy drag on the opposing tube's output, as the current from one tube subtracts from the other's output.
   Here you can see the painful inefficiency of Class A operation: it is not like two burly guys breaking into a sweat trying to work together to lift a safe up a flight of stairs; no, it is more like two burly guys fighting it out in a tug of war contest, with the safe tied to the middle of the rope. Each pulls against the other's efforts. If they pull equally, no one wins and the rope remains stationary. If one lessens his pull, the rope moves toward the other. So too it is with Class A operation: one device, whether it be transistor, MOSFET, pentode, or triode, pulls against the other and it is the difference between these two amounts of current flow that equals the amount that is delivered into the load.
   For example, if we start with two devices in the output stage and the idle current is set to 2 amps, then the load has no difference to absorb. If one device increases its conduction and is drawing 3 amps of current, while the second conducts less and is drawing 1 amp of current, then the load is absorbing the 2 amp difference. When the ratio is almost 4 amps for one device over almost no amps (1 amps) for the other, the amplifier is at its Class A limit, no more watts can be put out without moving the amplifier into Class AB operation.
   What if there is no load present? In this case the current flow through one device
must match

the other, as there is but one current path present in the output stage. No output current, no load, no watts. What if the output is shorted to ground? The output of the amplifier is fixed at zero volts, but the current flowing into the ground will obey the same math: device 1's current flow minus device 2's current flow. At idle each device conducts 2 amps, thus 2-2 = 0. At one extreme, 4-0 = 4 and at the other, 0-4 = -4. It would not make much sense to talk about the power output in this case, as any amount of current into 0 ohms equals 0 watts, which is a good thing, as otherwise the juice delivered to your house would be completely dissipated by the power company's cables. For DC current,
   Power = Current² x Resistance.
For RMS watts with sine waves, 
   Power = Current² x Resistance/2.
By the way, the 4 amps of current delivered into an 8 ohm load equals 64 watts of power.
  You can see immediately a problem with the claim that an amplifier with an idle current of about 0.5 amps being able to put out 60 watts of Class A power into an 8 ohm load. Each device (or set of devices) can only give up its half of an amp of current, which when met with a one half amp increase in current from the opposing device, only equals 1 amp of current, or in terms of watts, only 4 RMS watts of power into an 8 ohm load. The only amplifier that can meet the 60 RMS Class A watts specification is one that has an idle current of at least 2 amps. If the amplifier is tube based and it uses 135 volts per power supply leg, then the total plate dissipation for the 60 watt Class A amplifier would be 540 watts! This is just plate dissipation and the heater string would require over one hundred watts as well. Class A operation is not easy. In fact, as every electronic textbook in the world points out, the theoretical limit on any Class A  amplifier is 50%, with perfect, loss-less output devices and tube are much less than perfect and very loss-full.

pg. 19

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